Let $X$ be an affine holomorphic symplectic variety of dimension $2n$, with the associated Poisson bracket { , }. Let's say it's an integrable system when there are $n$ algebraically independent holomorphic functions $I_i$ ($i=1,\ldots,n$) on $X$ such that they Poisson-commute: $\{I_i,I_j\}=0$.

Pick a simple Lie algebra $\mathfrak{g}$, and pick a coadjoint orbit $O\subset \mathfrak{g}^*$. As is well-known, this is naturally a holomorphic symplectic variety: $x,y\in \mathfrak{g}$ gives a function on $O$, and its Poisson bracket is then given by the Lie bracket:

$\{x,y\}_{PB}=[x,y]$.

I wonder which coadjoint orbit is integrable, in the sense given above.

For example, any regular orbit $O$ of $\mathfrak{g}=\mathfrak{gl}(n)$ is integrable: let $X:\mathfrak{g}^*\to\mathfrak{g}$ be the identification via the invariant inner product, choose a generic element $a\in \mathfrak{g}$, and consider a function on $O$ given by

$P_k(t)= \mathrm{tr} (X+ta)^k$

where $t$ is a complex number. (I'm sorry for a slightly confusing notation, but $a$ here is a constant function on $O$ taking value in $\mathfrak{g}$.)

After some manipulation, $P_k(t)$ and $P_{k'}(t')$ are seen to Poisson-commute. Therefore, coefficients $P_{k,i}$ of $t^i$ of $P_k(t)$ all Poisson-commute. Note that $P_{k,k}$ is a constant function on $O$, because $O$ is a coadjoint orbit. Then there are in total $1+2+\cdots+(n-1) = (\dim \mathfrak{gl}(n)-\mathrm{rank} \mathfrak{gl}(n))/2$ independent commuting operators.

Let me conjecture that all coadjoint orbits are integrable.

Another large class of affine holomorphic symplectic varieties are Nakajima's quiver varieties, which include all the coadjoint orbits of $\mathfrak{gl}(n)$. A similar question can be asked: which quiver variety is integrable in this sense?

Update

Thanks to the answers so far, I could track down references, see e.g. the end of Sec. 4 of this paper, showing that any coadjoint orbit of real compact Lie algebras is integrable. I guess the proof should carry over to the semisimple orbits (and presumably nilpotent Richardson orbits) of complex semisimple Lie algebras, in the holomorphic sense. So the question now is: how about nilpotent orbits in $\mathfrak{g}_{\mathbb{C}}$?

Update 2

Indeed, A. Joseph says in this article that he essentially proved that Richardson orbits are integrable in this article (in the setting of enveloping algebras, not the associated graded.) It would be interesting e.g. non-special orbits are not integrable... but I have no idea.

Some comments. Actually Gelfand-Ceitlin system can be obtained as a limit of $tr(X+ta)^k$ taking $a=diag(u^n, u^{n-1}...u)$ and u-> 0. This some idea result by E. Vinberg, On some commutative subalgebras in universal enveloping algebra, Izv. AN USSR, Ser. Mat., 1990, vol. 54 N 1 pp 3-25
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Alexander ChervovJan 16 '12 at 19:11

From the Gaudin's point view matrix "a" is "constant magnetic field" acting on ALL spins. There is GL(n)-GL(m) duality (Howe's duality) which actually interchanges the matrix "a" and "Gaudin's marked point $z_i$" - it is presumable related to duality between monopoles - that "n-monopoles of charge m" is the same as "m-monom with charge n". Being more precise for Gaudin such duality is present only when we have n representations $S(C^m)$, so we have $S(C^m)\otimes ... S(C^m)= S(C^n\otimes C^m)= S(C^m)\otimes ... S(C^m) $ ... see next comment
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Alexander ChervovJan 16 '12 at 19:19

It is rather easy to check that GL(n) -Gaudin and GL(m) Gaudin defines the same algebra of commutative hamiltonians it is basically matrix identity $det(t+ A + X(z-B)^{-1} P ) = det( z-B + X(t+A)^{-1}P)$ - so you see that role of matrices "A" and "B" interchanged. This actually can be quantized - and this is Capelli identity arxiv.org/abs/math/0610799 A generalization of the Capelli identity E. Mukhin, V. Tarasov, A. Varchenko
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Alexander ChervovJan 16 '12 at 19:23

Our point of view that we may take Vinberg's limit $a=diag(u^n,u^{n−1}...u)$ and $u-> 0$ on both sides of GLn-GLm duality and we get Gelfand-Ceitlin dual to "bending flows", and it is more or less obvious that bending flows are the same as Jacys-Murphy elements.
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Alexander ChervovJan 16 '12 at 19:32

3 Answers
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Here is answer (YES) from Alexey Bolsinov who is one the main experts in these questions.

"The answer is YES

There is a very general construction allowing to construct an integrable system on more or less any coadjoint orbit for an arbitrary Lie algebra (non necessarily semisimple). This is a recent paper by Vinberg and Yakimova available in arxiv

Thank you; this paper springerlink.com/content/kfvevf4a13g0t3pp seems to explicitly construct involutive Hamiltonians on $T^*(G/P)$ (I'm confused if I'm talking about hol. sympl. spaces or just sympl. spaces here...) Do you have any comments on nilpotent orbits?
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Yuji TachikawaJan 12 '12 at 11:21

Interesting fact. Can we understand it somehow ? Toy model for such considerations - take sl_2 = C^3, regular orbit - is qaudric defined by Casimir = Const. Let me point out that Orbit - affine complex manifold and TP - is NOT, since contains a projective P. So they cannot be isomorphic as complex manifolds. But according to what I heard Orbit - is deformation of TP.
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Alexander ChervovJan 12 '12 at 12:54

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@Alexander: You're right.The orbit is an affine bundle over $G/P$, not a vector bundle. In the toy example you map the orbit to $\mathbb{P}^1$ by sending a (regular) matrix to the flag determined by an eigenvector. The affine bundle doesn't have a holomorphic section, but has a smooth one.
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Peter DalakovJan 12 '12 at 15:37

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@Alexander: I should've been more explicit about the section, sorry. Use the hermitian metric on $\mathbb{C}^2$: given a point $p\in\mathbb{P}^1$ take a unit vector $v_1\in p$. Then take an orthogonal unit vector, say $v_2$. Let $P$ be the matrix with columns $v_i$. The (smooth) section is $p\mapsto PDP^{-1}$, where $D=\textrm{diag}(\sqrt{\lambda},-\sqrt{\lambda})$. Here $\lambda\in \mathbb{C}^\times$ is the value of the determinant, which is fixed on the orbit.
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Peter DalakovJan 12 '12 at 16:37

@Peter Thank You very much ! Is it possible to see the result in this toy model ?
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Alexander ChervovJan 12 '12 at 20:32

My feeling is that for semi-simple g (at least classical g) this should be known to be true. But I cannot provide a reference now.

I am not sure about details, but the similar sounding conjecture sometimes associated with names of A.S. Mishchenko and A.T. Fomenko (early 80-ies ???). As far as I understand their conjecture is that any (not only semi-simple) g has maximal Poisson-commutative subalgebra in S(g). I am not sure did they conjecture that we can find int. sys. on any orbit. May be not explicitly. There is a bulk of works on this conjecture. Some names - A. Bolsinov, Trofimov, Fomenko himself wrote books on it, many of his students worked on it. I would suggest to look at

As far as I understand they rise question not only for g, but also for any affine Poisson manifold (I am not sure this work or not - but Yakimova surely discussed it).

About quivers - Nekrasov's old paper contains some examples of int.sys. on quivers arxiv.org/abs/hep-th/9503157 . My feeling was that one can see (at least some of) quivers as moduli spaces on vector bundles on very degenerated curves and so these are in a sense Hitchin's system.

Thank you very much as always. Indeed, the statement contained in the reference you gave me settles my question positively for regular orbits. Do you know any paper where non-regular orbits are studied?
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Yuji TachikawaJan 12 '12 at 8:06

Aaah... so this is the Gelfand-Zeitlin system (for $gl(n)$) I often hear about recently.
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Yuji TachikawaJan 12 '12 at 9:10

Thanks for thanks:). I'll try to e-mail to some colleagues in evening - the question must be discussed somewhere. I am also interested in it, but never had a motivations to ask... Since it is interesting not only for me - it gives some motivation.
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Alexander ChervovJan 12 '12 at 13:44